 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1/section-b/b6/b/solution

On the lower plane outside the boss, write

s

for distance from the boss centre along the plane, so

s > a

. Differentiate the potential normally before putting

z = 0

\frac{\partial Φ}{\partial z}_{z = 0} = E_{0} (1 - \frac{a ^{3}}{s ^{3}}) .

Thus the normal field is

E_{z} = - E_{0} (1 - a^{3} / s^{3})

and

σ_{plane} = - \frac{ε _{0} V}{d} (1 - \frac{a ^{3}}{s ^{3}}), s > a .

The source uses

r

for this distance on the plane. Near the rim it tends to zero, while far from the boss it recovers the uniform parallel-plate surface charge.

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